Question

6х-3х^2=0 3х^2-12х=0 7y^2+y=0 4y-y^2=0 х^4-17х^2+16=0 х^4+5^2-36=0 х^4-8^2+16=0 х^4+5x^2+9=0

Answer 1

1)
x(6-3x)=0
[x=0      [x=0
[-3x=-6  [x=2
2)
x(3x-12)=0
[x = 0      [x=0
[3x = 12  [x=4
3)
y(7y+1)=0
[y=0      [y = 0
[7y = -1 [y = -(1/7)
4)
y(4-y)=0
[y = 0
[y = 4

Answer 2

Let's solve each of these equations:

1. 6x - 3x^2 = 0: First, factor out the common term '3x' from both terms: 3x(2 - x) = 0 Now, set each factor equal to zero: 3x = 0 => x = 0 2 - x = 0 => x = 2

2. 3x^2 - 12x = 0: Factor out '3x' from both terms: 3x(x - 4) = 0 Set each factor equal to zero: 3x = 0 => x = 0 x - 4 = 0 => x = 4

3. 7y^2 + y = 0: Factor out 'y' from both terms: y(7y + 1) = 0 Set each factor equal to zero: y = 0 7y + 1 = 0 7y = -1 y = -1/7

4. 4y - y^2 = 0: Factor out 'y' from both terms: y(4 - y) = 0 Set each factor equal to zero: y = 0 4 - y = 0 -y = -4 y = 4

5. x^4 - 17x^2 + 16 = 0: This equation can be factored as a quadratic in terms of x^2: (x^2 - 16)(x^2 - 1) = 0 Now, solve for x: x^2 - 16 = 0 x^2 = 16 x = ±4

   x^2 - 1 = 0 x^2 = 1 x = ±1

So, the solutions for this equation are x = 4, x = -4, x = 1, and x = -1.

6. x^4 + 5x^2 + 9 = 0: This is a quartic equation, but it doesn't factor easily. You can try to solve it using numerical methods or a computer algebra system. The solutions are not simple rational numbers.

If you need more details or assistance with any of these equations, feel free to ask.